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Combination Sum

2024-07-17 22:30:07   220人阅读

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

  • All numbers (including target) will be positive integers.
  • Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
  • The solution set must not contain duplicate combinations.

 

For example, given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

分析：这道题类似combinaions和subsets，不同的是，在本题的可行解中可以存在不限数目的相同的数，同样我们用改良的dfs方法求解。因为此题要求每个combination中元素必须是非递减排列的，所以我们先将元素按升序排序。此外由于元素全为正数，当当前path的sum大于target时，即可停止该path的搜素。代码如下：

 1 class Solution { 2 public: 3     vector<vector<int> > combinationSum(vector<int> &candidates, int target) { 4         vector<vector<int>> res; 5         vector<int> path; 6         sort(candidates.begin(),candidates.end()); 7         dfs(res,path,candidates,0,0,target); 8         return res; 9     }10     void dfs(vector<vector<int>> &res, vector<int> & path, vector<int> & c, int start, int sum, int target){11         for(int i = start; i < c.size(); i++){12             path.push_back(c[i]);13             if(sum + c[i] == target)res.push_back(path);14             else if(sum + c[i] < target){15                 dfs(res,path,c,i,sum+c[i],target);16             }17             path.pop_back();18         }19     }20 };

 


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